3.132 \(\int \coth ^4(c+d x) (a+b \text {sech}^2(c+d x))^3 \, dx\)

Optimal. Leaf size=60 \[ a^3 x-\frac {(a+b)^3 \coth ^3(c+d x)}{3 d}-\frac {(a-2 b) (a+b)^2 \coth (c+d x)}{d}+\frac {b^3 \tanh (c+d x)}{d} \]

[Out]

a^3*x-(a-2*b)*(a+b)^2*coth(d*x+c)/d-1/3*(a+b)^3*coth(d*x+c)^3/d+b^3*tanh(d*x+c)/d

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Rubi [A]  time = 0.10, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4141, 1802, 207} \[ a^3 x-\frac {(a+b)^3 \coth ^3(c+d x)}{3 d}-\frac {(a-2 b) (a+b)^2 \coth (c+d x)}{d}+\frac {b^3 \tanh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Coth[c + d*x]^4*(a + b*Sech[c + d*x]^2)^3,x]

[Out]

a^3*x - ((a - 2*b)*(a + b)^2*Coth[c + d*x])/d - ((a + b)^3*Coth[c + d*x]^3)/(3*d) + (b^3*Tanh[c + d*x])/d

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 4141

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[((d*ff*x)^m*(a + b*(1 + ff^2*x^2)^(n/2))^p)/(1 + ff^
2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && IntegerQ[n/2] && (IntegerQ[m/2] ||
EqQ[n, 2])

Rubi steps

\begin {align*} \int \coth ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \left (1-x^2\right )\right )^3}{x^4 \left (1-x^2\right )} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (b^3+\frac {(a+b)^3}{x^4}+\frac {(a-2 b) (a+b)^2}{x^2}-\frac {a^3}{-1+x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{d}\\ &=-\frac {(a-2 b) (a+b)^2 \coth (c+d x)}{d}-\frac {(a+b)^3 \coth ^3(c+d x)}{3 d}+\frac {b^3 \tanh (c+d x)}{d}-\frac {a^3 \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=a^3 x-\frac {(a-2 b) (a+b)^2 \coth (c+d x)}{d}-\frac {(a+b)^3 \coth ^3(c+d x)}{3 d}+\frac {b^3 \tanh (c+d x)}{d}\\ \end {align*}

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Mathematica [B]  time = 1.73, size = 343, normalized size = 5.72 \[ \frac {\text {csch}(c) \text {sech}(c) \text {csch}^3(c+d x) \text {sech}(c+d x) \left (-16 a^3 \sinh (2 (c+d x))+8 a^3 \sinh (4 (c+d x))+8 a^3 \sinh (2 (c+2 d x))-12 a^3 \sinh (4 c+2 d x)-3 a^3 d x \cosh (2 (c+2 d x))-6 a^3 d x \cosh (4 c+2 d x)+3 a^3 d x \cosh (6 c+4 d x)-4 a^3 \sinh (2 d x)+6 a^3 d x \cosh (2 d x)-12 a^2 b \sinh (2 (c+d x))+6 a^2 b \sinh (4 (c+d x))+6 a^2 b \sinh (2 (c+2 d x))-18 a^2 b \sinh (4 c+2 d x)-18 a^2 b \sinh (2 c)+6 a^2 b \sinh (2 d x)+24 a b^2 \sinh (2 (c+d x))-12 a b^2 \sinh (4 (c+d x))-12 a b^2 \sinh (2 (c+2 d x))-36 a b^2 \sinh (2 c)+24 a b^2 \sinh (2 d x)+8 b^3 \sinh (2 (c+d x))-4 b^3 \sinh (4 (c+d x))-16 b^3 \sinh (2 (c+2 d x))+32 b^3 \sinh (2 d x)\right )}{96 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[c + d*x]^4*(a + b*Sech[c + d*x]^2)^3,x]

[Out]

(Csch[c]*Csch[c + d*x]^3*Sech[c]*Sech[c + d*x]*(6*a^3*d*x*Cosh[2*d*x] - 3*a^3*d*x*Cosh[2*(c + 2*d*x)] - 6*a^3*
d*x*Cosh[4*c + 2*d*x] + 3*a^3*d*x*Cosh[6*c + 4*d*x] - 18*a^2*b*Sinh[2*c] - 36*a*b^2*Sinh[2*c] - 4*a^3*Sinh[2*d
*x] + 6*a^2*b*Sinh[2*d*x] + 24*a*b^2*Sinh[2*d*x] + 32*b^3*Sinh[2*d*x] - 16*a^3*Sinh[2*(c + d*x)] - 12*a^2*b*Si
nh[2*(c + d*x)] + 24*a*b^2*Sinh[2*(c + d*x)] + 8*b^3*Sinh[2*(c + d*x)] + 8*a^3*Sinh[4*(c + d*x)] + 6*a^2*b*Sin
h[4*(c + d*x)] - 12*a*b^2*Sinh[4*(c + d*x)] - 4*b^3*Sinh[4*(c + d*x)] + 8*a^3*Sinh[2*(c + 2*d*x)] + 6*a^2*b*Si
nh[2*(c + 2*d*x)] - 12*a*b^2*Sinh[2*(c + 2*d*x)] - 16*b^3*Sinh[2*(c + 2*d*x)] - 12*a^3*Sinh[4*c + 2*d*x] - 18*
a^2*b*Sinh[4*c + 2*d*x]))/(96*d)

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fricas [B]  time = 0.40, size = 354, normalized size = 5.90 \[ -\frac {{\left (4 \, a^{3} + 3 \, a^{2} b - 6 \, a b^{2} - 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{4} - 4 \, {\left (3 \, a^{3} d x + 4 \, a^{3} + 3 \, a^{2} b - 6 \, a b^{2} - 8 \, b^{3}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (4 \, a^{3} + 3 \, a^{2} b - 6 \, a b^{2} - 8 \, b^{3}\right )} \sinh \left (d x + c\right )^{4} + 9 \, a^{2} b + 18 \, a b^{2} + 4 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + 4 \, b^{3}\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (2 \, a^{3} + 6 \, a^{2} b + 6 \, a b^{2} + 8 \, b^{3} + 3 \, {\left (4 \, a^{3} + 3 \, a^{2} b - 6 \, a b^{2} - 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{2} - 4 \, {\left ({\left (3 \, a^{3} d x + 4 \, a^{3} + 3 \, a^{2} b - 6 \, a b^{2} - 8 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} - {\left (3 \, a^{3} d x + 4 \, a^{3} + 3 \, a^{2} b - 6 \, a b^{2} - 8 \, b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{12 \, {\left (d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (d \cosh \left (d x + c\right )^{3} - d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^4*(a+b*sech(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

-1/12*((4*a^3 + 3*a^2*b - 6*a*b^2 - 8*b^3)*cosh(d*x + c)^4 - 4*(3*a^3*d*x + 4*a^3 + 3*a^2*b - 6*a*b^2 - 8*b^3)
*cosh(d*x + c)*sinh(d*x + c)^3 + (4*a^3 + 3*a^2*b - 6*a*b^2 - 8*b^3)*sinh(d*x + c)^4 + 9*a^2*b + 18*a*b^2 + 4*
(a^3 + 3*a^2*b + 3*a*b^2 + 4*b^3)*cosh(d*x + c)^2 + 2*(2*a^3 + 6*a^2*b + 6*a*b^2 + 8*b^3 + 3*(4*a^3 + 3*a^2*b
- 6*a*b^2 - 8*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 - 4*((3*a^3*d*x + 4*a^3 + 3*a^2*b - 6*a*b^2 - 8*b^3)*cosh(
d*x + c)^3 - (3*a^3*d*x + 4*a^3 + 3*a^2*b - 6*a*b^2 - 8*b^3)*cosh(d*x + c))*sinh(d*x + c))/(d*cosh(d*x + c)*si
nh(d*x + c)^3 + (d*cosh(d*x + c)^3 - d*cosh(d*x + c))*sinh(d*x + c))

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giac [B]  time = 0.34, size = 155, normalized size = 2.58 \[ \frac {3 \, a^{3} d x - \frac {6 \, b^{3}}{e^{\left (2 \, d x + 2 \, c\right )} + 1} - \frac {2 \, {\left (6 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} + 9 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} - 3 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} - 6 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 18 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 12 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 4 \, a^{3} + 3 \, a^{2} b - 6 \, a b^{2} - 5 \, b^{3}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^4*(a+b*sech(d*x+c)^2)^3,x, algorithm="giac")

[Out]

1/3*(3*a^3*d*x - 6*b^3/(e^(2*d*x + 2*c) + 1) - 2*(6*a^3*e^(4*d*x + 4*c) + 9*a^2*b*e^(4*d*x + 4*c) - 3*b^3*e^(4
*d*x + 4*c) - 6*a^3*e^(2*d*x + 2*c) + 18*a*b^2*e^(2*d*x + 2*c) + 12*b^3*e^(2*d*x + 2*c) + 4*a^3 + 3*a^2*b - 6*
a*b^2 - 5*b^3)/(e^(2*d*x + 2*c) - 1)^3)/d

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maple [B]  time = 0.48, size = 149, normalized size = 2.48 \[ \frac {a^{3} \left (d x +c -\coth \left (d x +c \right )-\frac {\left (\coth ^{3}\left (d x +c \right )\right )}{3}\right )+3 a^{2} b \left (-\frac {\cosh \left (d x +c \right )}{2 \sinh \left (d x +c \right )^{3}}-\frac {\left (\frac {2}{3}-\frac {\mathrm {csch}\left (d x +c \right )^{2}}{3}\right ) \coth \left (d x +c \right )}{2}\right )+3 a \,b^{2} \left (\frac {2}{3}-\frac {\mathrm {csch}\left (d x +c \right )^{2}}{3}\right ) \coth \left (d x +c \right )+b^{3} \left (-\frac {1}{3 \sinh \left (d x +c \right )^{3} \cosh \left (d x +c \right )}+\frac {4}{3 \sinh \left (d x +c \right ) \cosh \left (d x +c \right )}+\frac {8 \tanh \left (d x +c \right )}{3}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(d*x+c)^4*(a+b*sech(d*x+c)^2)^3,x)

[Out]

1/d*(a^3*(d*x+c-coth(d*x+c)-1/3*coth(d*x+c)^3)+3*a^2*b*(-1/2/sinh(d*x+c)^3*cosh(d*x+c)-1/2*(2/3-1/3*csch(d*x+c
)^2)*coth(d*x+c))+3*a*b^2*(2/3-1/3*csch(d*x+c)^2)*coth(d*x+c)+b^3*(-1/3/sinh(d*x+c)^3/cosh(d*x+c)+4/3/sinh(d*x
+c)/cosh(d*x+c)+8/3*tanh(d*x+c)))

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maxima [B]  time = 0.52, size = 366, normalized size = 6.10 \[ \frac {1}{3} \, a^{3} {\left (3 \, x + \frac {3 \, c}{d} - \frac {4 \, {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} - 2\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}}\right )} + 4 \, a b^{2} {\left (\frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}} - \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}}\right )} + \frac {16}{3} \, b^{3} {\left (\frac {2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} - 2 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )} - 1\right )}} - \frac {1}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} - 2 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )} - 1\right )}}\right )} + 2 \, a^{2} b {\left (\frac {3 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}} + \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^4*(a+b*sech(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

1/3*a^3*(3*x + 3*c/d - 4*(3*e^(-2*d*x - 2*c) - 3*e^(-4*d*x - 4*c) - 2)/(d*(3*e^(-2*d*x - 2*c) - 3*e^(-4*d*x -
4*c) + e^(-6*d*x - 6*c) - 1))) + 4*a*b^2*(3*e^(-2*d*x - 2*c)/(d*(3*e^(-2*d*x - 2*c) - 3*e^(-4*d*x - 4*c) + e^(
-6*d*x - 6*c) - 1)) - 1/(d*(3*e^(-2*d*x - 2*c) - 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) - 1))) + 16/3*b^3*(2*e^
(-2*d*x - 2*c)/(d*(2*e^(-2*d*x - 2*c) - 2*e^(-6*d*x - 6*c) + e^(-8*d*x - 8*c) - 1)) - 1/(d*(2*e^(-2*d*x - 2*c)
- 2*e^(-6*d*x - 6*c) + e^(-8*d*x - 8*c) - 1))) + 2*a^2*b*(3*e^(-4*d*x - 4*c)/(d*(3*e^(-2*d*x - 2*c) - 3*e^(-4
*d*x - 4*c) + e^(-6*d*x - 6*c) - 1)) + 1/(d*(3*e^(-2*d*x - 2*c) - 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) - 1)))

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mupad [B]  time = 1.48, size = 260, normalized size = 4.33 \[ a^3\,x-\frac {\frac {2\,\left (a^2\,b+2\,a\,b^2+b^3\right )}{d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (2\,a^3+3\,a^2\,b-b^3\right )}{3\,d}}{{\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1}-\frac {\frac {2\,\left (2\,a^3+3\,a^2\,b-b^3\right )}{3\,d}+\frac {2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (2\,a^3+3\,a^2\,b-b^3\right )}{3\,d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^2\,b+2\,a\,b^2+b^3\right )}{d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1}-\frac {2\,b^3}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {2\,\left (2\,a^3+3\,a^2\,b-b^3\right )}{3\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(c + d*x)^4*(a + b/cosh(c + d*x)^2)^3,x)

[Out]

a^3*x - ((2*(2*a*b^2 + a^2*b + b^3))/d + (2*exp(2*c + 2*d*x)*(3*a^2*b + 2*a^3 - b^3))/(3*d))/(exp(4*c + 4*d*x)
- 2*exp(2*c + 2*d*x) + 1) - ((2*(3*a^2*b + 2*a^3 - b^3))/(3*d) + (2*exp(4*c + 4*d*x)*(3*a^2*b + 2*a^3 - b^3))
/(3*d) + (4*exp(2*c + 2*d*x)*(2*a*b^2 + a^2*b + b^3))/d)/(3*exp(2*c + 2*d*x) - 3*exp(4*c + 4*d*x) + exp(6*c +
6*d*x) - 1) - (2*b^3)/(d*(exp(2*c + 2*d*x) + 1)) - (2*(3*a^2*b + 2*a^3 - b^3))/(3*d*(exp(2*c + 2*d*x) - 1))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)**4*(a+b*sech(d*x+c)**2)**3,x)

[Out]

Timed out

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